AP Notes Task Answers 9 UNITS

UNIT 00 – AP Chemistry Preamble

0.1

1. (a) 2.45 x 10⁴

(b) 3.56 x 10²

(d) 2.22 x 10^-1

(e) 1.22 x 10⁴

2. (a) 4200

(b) 0.000215

(d) 922000

(e) 957

0.2

1. (a) 0.0186

(b) 0.852

(d) 74.97

(e) 15.4195

2. (a) 3 miles = 4828.03 m

(b) 25 lbs = 11.3398 kg

0.3

1. (a) -260

(b) 1663

2. When determining change, one degree Celsius is the same as (equivalent to) one Kelvin.

0.4

1. (a) 4

(b) 2

(d) 2

(e) 2

(f) 4

2. (a) 809

(b) 40

(d) 67.64

(e) 12.7

(f) 3

0.5

1. (a) Y

(b) Y

0.6

1. (a) p = 82, e = 82, n = 128

(b) p = 16, e = 16, n =18

2. (a) 20

(b) 10

0.7

1. (a) Sodium chloride

(b) Strontium oxide

(d) Barium chloride

(e) Potassium oxide

(f) Copper(II) oxide

(g) Copper(I) oxide

2. (a) Mg₃N₂

(b) BaBr₂

(d) KI

(e) LiCl

(f) NaF

(g) SnBr₄

0.8

1. (a) NH₄NO₃

(b) CuBr₂

(d) Zn(HSO₄)₂

(e) Al₂(SO₄)₃

(f) NaClO₄

(g) Cu(IO₂)₂

2.(a) Sodium nitrate

(b) Potassium manganate(VII) (potassium permanganate)

(d) Copper(II) sulfate

(e) Copper(I) sulfate

(f) Potassium nitrite

(g) Lithium perchlorate

0.9

1. (a) N₂O₄

(b) PCl₅

(d) NO₂

(e) H₂O

2. (a) Dinitrogen pentoxide

(b) Phosphorus trichloride

(d) Water (dihydrogen monoxide)

(e) Dichlorine monoxide

UNIT 01 – Atoms, Elements and Periodicity

1.1

1. (a) FeO

(b) Fe₂O₃

2. CH, CH₂, CH₃O, C₃H₆O₂ – all cannot be simplified further.

3. They are not the same sample, not all can be pure.

1.2

1. CH

2. C₆H₆

3. Yes, BUT also consistent with an impurity that has any amount of C that is less than 92.31 since the % of C has gone down.

1.3

1. 58.5 g

2. 3.5 moles

3. 0.00234 moles

4. 0.523 moles

5. Answers here are the number of moles multiplied by Avogadro’s number (6.02 x 10²³), remembering that for example, NaCl has both Na⁺ ions and Cl^– ions and that CuBr₂ contains twice as many Br^– ions as Cu²⁺ ions, etc.

1.4

1. (a) No, different elements

(b) Yes, same element with different numbers of neutrons therefore different masses.

1.5

1. 20.18

2. 85.58

3. 69.76

1.6

1. 35.5; Cl³⁷ = 25%, Cl³⁵ = 75%

2. 5 peaks. Two of the same height at 79 and 81, and three at m/z values of 158, 160 and 162, with the 160 peak twice the size of the 158 and 162 which would be identical. Think of the bromine diatomic molecule as being made up of Br atom “x”, and Br atom “y”.

Combo a. x = 79 and y = 79, when added = 158

Combo b. x = 81 and y = 79, when added = 160

Combo c. x = 79 and y = 81, when added = 160

Combo d. x = 81 and y = 81, when added = 162

160 is twice as likely as 158 and 162, so has a peak twice the height.

3. 47.92

4.(a) That there are three isotopes of Mg

(b) (i) That it is the isotope with the greatest abundance

(ii) 24.3

1.7

1.8

1. (a) [Ar] 4s¹ 3d¹⁰

(b) [Ar] 4s² 3d⁷

(d) [He] 2s² 2p²

(e) [Ne] 3s² 3p⁶

(f) [Ar] 4s² 3d¹⁰ 4p¹

2. (a) [He] 2s² 2p⁶ OR [Ne]

(b) [Ne] 3s² 3p⁶ OR [Ar] OR [Ar] 4s⁰

(d) [He] 2s² 2p⁶ OR [Ne]

(e) [He] 2s² 2p⁶ OR [Ne]

1.9

1.10

1. (a) There are 4 occupied sub-shells

(b) 1s² 2s² 2p⁶ 3s¹ in decreasing order of peak energy (i.e., the 1s² electrons are the most difficult to remove)

2. (a) Three

(b) 1s²

3. Aluminum. It is the first element to have electrons in the 3p sub-shell. The 3p sub-shell is the fifth sub-shell to be filled

1.11

1. (a) Decreases. Greater shielding outweighs greater number of protons
(b) Increases. Same shielding, more protons

2. (a) Group 1
(b) Group 2
(c) It only has three electrons

3. Energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of singly charge ions

Number of protons and the amount of shielding (distance form the nucleus).

4. (a) RO
(b) Less mutual repulsion of electrons but same number of protons

5. Increases. Same shielding, more protons

6. Zero shielding, electrons directly adjacent to the nucleus

1.12

Isoelectronic, BUT different numbers of protons. Largest number of protons = greatest attraction = smallest species

UNIT 02 – Chemical Bonding I

2.1

Na⁺, Cl^–; NaCl

Ca²⁺, Cl^–; CaCl₂

Fe³⁺, Br^–; FeBr₃

Na⁺, O^2-; Na₂O

NaCl. The smaller radius of the chloride ion when compared to the bromide ion allows the ions in NaCl to get closer to one another than in NaBr, and therefore the Coulombic attraction is greater.

2.2

2.3

UNIT 03 – Chemical Bonding II & Gases

3.1

1. (a) The LDF’s are approximately the same.

(b) Butan-1-ol has additional hydrogen bonding as an IMF that dramatically increases the attractions between molecules of the alcohol, thus significantly raising the boiling point when compared to pentane

2. Both molecules have LDF’s, but HCl also has dipole-dipole interactions

Despite dipole-dipole interactions generally being stronger than LDF’s, in instances where the molecules are large (larger surface area and more electrons), the collective interactions created by many LDF’s can outweigh the stronger dipole-dipole interactions to give a greater attraction between the molecules. In this case, the larger Cl₂ molecules have LDF’s that are collectively stronger than the LDF’s and dipole-dipole interactions in the much smaller, HCl

3. Increase in boiling points since LDF’s increase with surface area and number of electrons and polarizability.

3.2

A: Ionic – poor conductor when solid because ions cannot move. High m.p., means strong, ionic bonds. Will conduct when liquid when the ions are free to move.

B: Metallic – high m.p., always a good conductor with free moving electrons.

C: Molecular – never a good conductor (no charged particles to act as charge carriers. Low m.p. due to weak IMF’s.

3.3

1. 0.864 atm

2. 1.09 atm

3.4

1. 0.272 L

2. 422 mL

3.5

1. 10.4 L

2. 288 K

3.6

1. 8.21 L

2. 37.4 L

3.7

1. 318 K

2. 247 K

3.8

1. 0.435 mols

2. 266.4 g mol^-1

3.9

1. 0.0255 mols

2. 717.6 kPa

Apply P1V1 = P2V2, Boyles law for EACH gas, then add the new P’s (Dalton’s).

3.10

1. (a) 1363 ms^-1

(b) 482 ms^-1

2. (a) 1419 ms^-1

(b) 502 ms^-1

3. u_rms is lower at lower temperatures and lower with larger molar masses

3.11

1. 0.192 L (192 mL)

2. 0.192 L (192 mL)

3. 1.75 L

4. Add approx. 250 mL of distilled/deionized water to a 500 mL volumetric flask. Slowly and carefully add 192 mL of nitric acid using a buret. Make up to the mark with distilled/deionized water. Wear safety goggles, chemical resistant gloves and lab coat or apron at all times, and only ever add small amounts of concentrated acid to large volumes of water. Diluting concentrated acid can be a highly exothermic process, and safe dilution is essential. The use of accurately graduated glassware such as burets, pipets and volumetric flasks, ensure the accuracy of the dilution

UNIT 04 – Chemical Reactions

4.1

1. (a) H₂ + CuO → Cu + H₂O

(b) 2C + O₂→ 2CO

2. (a) Ca + 2H₂O → H₂+ Ca(OH)₂

(b) 2Cu + O₂→ 2CuO

(d) Fe + 2HCl → FeCl₂+ H₂

(e) 2Fe + 3Br₂→ 2FeBr₃

(f) C₄H₈ + 6O₂→ 4CO₂+ 4H₂O

(g) Na₂CO₃ + 2HI → 2NaI + CO₂+ H₂O

(h) CuCO₃ → CuO + CO₂

(i) 2Pb(NO₃)₂ → 2PbO + 4NO₂+ O₂

(j) H₂SO₄+ 2KOH → K₂SO₄+ 2H₂O

(k) 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂

(l) 4Al + 3O₂→ 2Al₂O₃

4.2

1. (a) Formation of steam (the reaction is sufficiently exothermic that it produces enough energy to convert the water produced to a gas)

(b) 2H₂ + O₂ → 2H₂O

(c) Reactants and products appear in the correct ratios (i.e., the equation is balanced according to the answer given in (b). Oxygen, hydrogen and water are all consistently represented by the black and white circles (i.e., the ratio of atoms in each case are consistent with the known formulas for the substances)

4.3

1. When water boils the process can be summarized as; H₂O_(l) → H₂O_(g). As such, no new substances are formed, there is only a change of state, and only intermolecular forces are broken

2. When water is broken down into its elements, the process can be summarized by the equation, 2H₂O → 2H₂ + O₂. As such, new substances are formed, and intra bonds (covalent) are broken

4.4

1. (a) 3.24 g

(b) 76.5 g

2. (a) 1:1

X + Br₂ → XBr₂

(b) Calcium (molar mass 40 gmol^-1) that is consistent with being a group 2 element, and forming a 2+ ion, and reacting in the ratio in the equation in (a)

3. 34.98 g

4. 1Cu²⁺:2I^–

5. 0.142 M

6. 0.981 g

7. 0.05 L or 50 mL

4.5

1. 27.7 g

2. 1.12 L

3. 83.7 gmol^-1

4.6

Chlorine is limiting.

1.42 moles of product (SCl₆) which is equal to 348 g

157 g of excess reactant (S) left over

4.7

88.9%

4.8

C₅H₆O₃

4.9

1. x = 6

2. Heating until the mass does not change anymore, i.e., heating until all of the water has been driven off and the salt is completely dehydrated (anhydrous)

4.10

1.14 M

4.11

2KOH_(aq) + H₂SO_4(aq) → K₂SO_4(aq) + 2H₂O_(l)

2K⁺_(aq) + 2OH^–_(aq) + 2H⁺_(aq) + SO₄^2-_(aq) → 2K⁺_(aq) + SO₄^2-_(aq) + 2H₂O_(l)

2H⁺_(aq) + 2OH^–_(aq) → 2H₂O_(l)

4.12

1. (a) +6

(b) -0.5

(d) +2

(e) +3

2. (a) Ca + S → CaS

Ca → Ca²⁺ + 2e^–

S + 2e^– → S^2-

(b) 2Al + 3Br₂ → 2AlBr₃

Al → Al³⁺ + 3e^–

Br₂ + 2e^– → 2Br^–

Al → Al³⁺ + 3e^–

O₂ + 4e^– → 2O^2-

3. (a) IO₃^– + 5e^– + 6H⁺ → ½I₂ + 3H₂O

5I^– → 2½I₂ + 5e^–

IO₃^– + 6H⁺ + 5I^– → 3H₂O + 3I₂

(b) I₂+ 2e^– → 2I^–

2S₂O₃^2- → S₄O₆^2- + 2e^–

2S₂O₃^2- + I₂ → S₄O₆^2- + 2I^–

2I^– → I₂ + 2e^–

Br₂ + 2KI → I₂ + 2KBr

(d) 2MnO₄^2- → 2MnO₄^– + 2e^–

MnO₄^2- + 2e^– + 4H⁺ → MnO₂ +2H₂O

3MnO₄^2- + 4H⁺ → 2MnO₄^–+ MnO₂ + 2H₂O

(e) NOT REDOX (all oxidation states the same) so no half reactions required

2CrO₄^2- + 2H⁺ → Cr₂O₇^2- + H₂O

4.13

1.(a) MgO

(b) It goes up, side oxygen is added.

2. (a) Decomposition.

(b) CaCO₃ → CaO + CO₂

Mass of CO₂ = 30.46 – 27.86 = 2.6 g

Moles of CaCO₃ = 5.90/100 = 0.0590

Moles of CO₂ = 2.6 /44 = 0.059

i.e., 1:1 ratio which agrees with the equation

4.14

0.0754 M

4.15

x = 12

Method

1. Calculate the moles of manganate(VII) by multiplying the concentration by the volume.

2. Realize (and THIS is really the quantum leap of intelligence) that in that huge, double salt, the thing that is getting oxidized is Fe²⁺, so therefore the missing half reaction is, Fe²⁺ → Fe³⁺ + e^–.

3. Combine the half reaction in 2. above with the manganate(VII) half reaction to get the reacting ratio (5Fe²⁺:1MnO₄^–) and therefore get the moles of Fe²⁺ in the 25.0 mL sample from the answer to 1., above.

4. Multiple the answer in 3. above by 10 to get moles of Fe²⁺ in 250 mL sample, and realize that the moles of Fe²⁺ also equals the moles of the WHOLE salt (since in 1 mole of the whole salt there is 1 mole of Fe²⁺, i.e. it’s a 1:1 ratio).

5. Calculate the molar mass of the whole salt by dividing grams (given in the Q) by moles calculated in 4. above.

6. Subtract all of the molar masses of the pieces of the salt that we know (N, H, Fe, S, O) to leave the mass of H₂O.

7. The answer to 6. above should be divisible nicely by 18, i.e., should give you an integer value for x.

4.16

(a) Mg²⁺_(aq) +2OH^–_(aq) → Mg(OH)_2(s)

(b) Ba²⁺_(aq) + S^2-_(aq) + Ni²⁺_(aq) + SO₄^2-_(aq) → NiS_(s) + BaSO_4(s)

(d) Al³⁺_(aq) +3OH^–_(aq) → Al(OH)_3(s)

4.17

(a) AgCl

(b) 0.002 mols

(d) 0.0709 g

(e) 39%

(f) The mass of precipitate would be too large, making the moles of AgCl too large, making the moles of chloride too large, making the mass of chloride too large, making the calculated % by mass too large.

UNIT 05 – Chemical Kinetics

5.1

(a) They are both catalysts since they both increase the rate of reaction, and at the end of the experiment they are regenerated, i.e., are unchanged

(b) In each case the H₃O⁺ and the OH^– are regenerated

5.2

1. The total mass of the chips was the same in experiment, since the total volume of gas produced in each experiment was the same. (The acid was in excess)

2. The small chip experiment was complete between 120 and 135 seconds. The large chip experiment was complete between 150 and 165 seconds. The reaction is complete when no more gas is produced which is illustrated on the graph when the line plateaus

3. mL/s (many other possible answer would work)

4. 2HCl_(aq) + CaCO_3(s) → CO_2(g) + H₂O_(l) + CaCl_2(aq)

5. When solids are present, one of the factors which affects the rate of a chemical reaction is the surface area of the solid. The reaction between the solution and the solid takes place at the surface of the solid, so a greater surface area offers more sites for collision, and therefore a faster reaction. Small chips have a larger surface area than the same mass of larger chips, so the reaction is faster with the smaller sized solid

6. Mass the chips using the balance, use the measuring cylinder to measure the volume of the acid, and then use the gas syringe to collect the carbon dioxide from the reaction occurring in a Erlenmeyer flask. See this for an example of the set up

7. Heat the acid solution to varying temperatures OR change the concentration of the acid to varying concentrations, each time keeping all other variables the same

5.3

V₂O₄ is the intermediate, V₂O₅ is the catalyst

5.4

1. Rate = k [X]²

Order w.r.t [Y] = 0

Total order = 2 + 0 = 2

k = 40000 M^-1 min^-1

2. Rate = k [A][B]

Total order = 1 + 1 = 2

k = 8000 M^-1 s^-1

5.5

2. Mechanism #2

3. X is an intermediate

5.6

1. 1st order. The graph of ln[A] versus time is a straight line, consistent with the integrated rate law for first order reactions

2. 7 minutes (approx. since ‘from graph’)

3. 1.01 x 10^-1 min^-1

Note: This value was calculated by taking the data across the full time range of 25 minutes. Other, very similar (but different) values can be calculated by using smaller time ranges

UNIT 06 – Chemical Thermodynamics I

6.1

1. (a) From propane to water
(b) 23.83 K
(c) Exothermic since energy is released during the combustion

6.2

1. (a) 7599 J
(b) 92.96 K

6.3

1. -124 kJ
2. -95 kJ
3. -1198 kJ
4. 0 (zero). Coincidentally, identical covalent bonds are broken and made so the endothermic (+ve) and exothermic (-ve) processes have the same magnitude and their sum = 0

6.4

1. Ionic bonds. Intermolecular, hydrogen bonds

2. Endothermic, bonds or forces being broken.

3. They are polar, exothermic

6.5

1. C_(graphite) + 1½H_2(g) + ½Br_2(l) → CH₃Br_(l)

2. 4C_(graphite) + 4H_2(g) + ½O_2(g) → CH₃COC₂H_5(l)

3. Na_(s) + ½N_2(g) + 1½O_2(g) → NaNO_3(s)

4. During the formation of an element, “from its elements”, i.e., from ITSELF, there is no change. For example; Na_(s) ==> Na_(s) represents no change in any aspect, of which enthalpy is just one

6.6

1. C₅H_12(l) + 😯_2(g)→ 5CO_2(g) + 6H₂O_(l)

2. CO_(g) + ½O_2(g) → CO_2(g)

3. H_2(g) + ½O_2(g) → H₂O_(l)

4. Al_(s) + ¾O_2(g) → ½Al₂O_3(s)

6.7

1. -84 kJ

2. -2005 kJ

3. (a) -852 kJ

(b) They are all zero

4. (a) S_(s) + O_2(g) → SO_2(g)

(b) -297 kJ

5. (a) -126 kJ

(b) -124 kJ

(c) Answer (a), since it is calculated from formation and combustion data that are specific to the compounds involved. In part (b), the bond enthalpies are average values and therefore not necessarily specific to the compounds involved

UNIT 07 – Chemical Equilibrium

7.1

1. Kc = 4.

All experiments yield the same value for K because each reaction is carried out at the same temperature

2. x = 0.845, so moles of acid = 0.155, alcohol = 1.155, ester = 0.845, water = 0.845

3. Kc = 13.1

4. Moles of HBr = 8.62

7.2

The relatively large amount of HI suggests that the equilibrium lies heavily to the right and side (products), and therefore the value of K will be relatively large

7.3

1. (a) Kp = 3.05

(b) The expression Kp = leads to a math equation that is beyond the scope of AP chemistry, but can be solved. Here it is simply important to see how the expression reached. (The actual value of x = 0.58 (or 58% dissociated) is consistent with a lower total pressure according to Le Châtelier’s principle)

2. (a) Kp =13.8

(b) Kp = 21.8

(c) The expression Kp = 2.11 x 10⁴ = leads to a math equation that is beyond the scope of AP, but can be solved. Here it is simply important to see how the expression reached. The actual value of x = 0.986 (or 98.6% dissociated) is consistent with the huge change in Kp at the higher temperature

3. 0.128

7.4

1. 1.72 x 10^-11

2. 2.36 x 10^-5

3. 0.0152 M, 4.75 g/L.

4. Precipitate forms since Q > Ksp and to reach equilibrium, the concentrations of the ions in solution must be reduced. That is achieved via the formation of a ppt.

7.5

1.(a) 1.26 x 10^-5

(b) 2.87 x 10^-8

2. Addition of sulfate ions from an external source makes Q much larger than K. In order to restore equilibrium, Q must decrease in order for the value of K to be re-stored, and the reaction must shift to the side where solid barium sulfate is.

7.6

1. 4.5

2. (a) 6.7 x 10^-6

(b) 1.2 x 10^-9

(c) The high concentration of hydroxide ions in a solution of pH 11, causes the equilibrium to be shifted backwards (Le Châtalier’s principle or Q v. K), and the solubility of the solid decreases. Essentially, this is a common ion problem.

UNIT 08 – Acid & Bases

8.1

1. Diagram B

2. Diagram C

3. Diagram A is a relatively dilute, weak acid; Diagram D is a relatively dilute, strong acid

8.2

1. 1.5

2. 12

3. 4.8 x 10^-5

4. 13.15

8.3

1. 1.82 x 10^-5

2. 4.86

3. Ethanoic acid since it has a larger Ka, meaning a greater degree of ionization

4. 2.37

5. 3.06 x 10^-5

6. 2.41 x 10^-10

8.4

1. 1.78 x 10^-5

2. 3.36

3. Methylamine since it has a larger Kb, meaning a greater degree of ionization

4. 11.45

5. 3.06 x 10^-5

6. 6.95 x 10^-10

8.5

1. 4.74

2. The best buffers are generally those with an approx. 1:1 ratio of acid and conjugate base (salt) components, since this allows them to absorb any additional acid and base equally well. Since the log of 1 is equal to 0, then the pH of the buffer is equal to its pKa

3. pH = pKa + log (deprotonated/protonated) = 7.6 + log (0.67/0.33) = 7.9

4. 4.62

5. 5.13

UNIT 09 – Chemical Thermodynamics II & Electrochemistry

9.1

1. (a) Negative

(b) Positive

2. (a) 400 K. Higher temperature means greater disorder (entropy)

(b) A soft metal will have flexible, metallic bonding that induces more disorder than a very rigid solid like diamond

(c) The gas at the lower pressure, since the particles have more space to move in and as such can be more disordered. A gas at a relatively high pressure, means that the particles are packed into a relatively small space with relatively little room to move around. This leads to a lower level of disorder, i.e., a lower entropy

9.2

(a) Zn_(s) | Zn²⁺_(aq) || 1/2 F_2(g) | F^–_(aq) | (Pt)
(b) Sn_(s) | Sn²⁺_(aq) || H⁺_(aq) | 1/2 H_2(g) | (Pt)
(c) Fe_(s) | Fe²⁺_(aq) || Cu²⁺_(aq) | Cu_(s) |
(d) (Pt) | 1/2 H_2(g) | H⁺_(aq) || Fe³⁺_(aq), Fe²⁺_(aq) | (Pt)